Q.74. Algebraic inequality

Question 74.
If a,b,c > 0 and ab + bc + ca = 1 - 2abc, then prove that  2(a + b + c) + 1 ≥ 32abc.

Answer 74.
AM ≥ HM
=> (ab + bc + ca + 2abc)/4 ≥ 4 / [1/ab + 1/bc + 1/ca + 1/2abc)] ... (1)

Now, ab + bc + ca + 2abc = 1
and 1/ab + 1/bc + 1/ca + 1/2abc = [2(a + b + c) + 1] / 2abc

Plugging in ( 1 ),
1/4 ≥ 4 / [2(a + b + c) + 1] / 2abc
=> 2(a + b + c ) + 1 ≥ 32abc.

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